# Question Regarding the Coordinate Independent Form of the Exterior Derivative

The coordinate version of the exterior derivative $d:\Omega^k(M)\to \Omega^{k+1}(M)$ of differential forms of a $C^\infty$ manifold $M$ can be expressed on a form $$\omega=fdx^1\wedge\cdots\wedge dx^k$$ as $$d\omega=\sum_{i=1}^n\frac{\partial f}{\partial x^i}dx^i\wedge dx^1\wedge\cdots \wedge dx^k$$ (for example). Alternatively, one can express this in a coordinate invariant form as $$d\omega(X_1,\ldots, X_{k+1})=\sum_{i=1}^{k+1}(-1)^{i+1}X_i(\omega(X_1,\ldots, \widehat{X_i},\ldots, X_{k+1}))$$ $$+\sum_{1\le i<j\le k+1}(-1)^{i+j}\omega([X_i,X_j],X_1,\ldots, \widehat{X_i},\ldots, \widehat{X_j},\ldots,X_{k+1}).$$ Here the $X_i\in \mathfrak{X}(M)$. I can show that these two operators are actually the same. That's not a problem. My confusion is in showing that the coordinate invariant form $d\omega$ is independent of extension $X_1,\ldots, X_{k+1}\in \mathfrak{X}(M)$. Indeed, the idea is that we define $d\omega_p$ on a $(k+1)-$tuple $$(v_1,\ldots, v_{k+1})\in \overbrace{T_pM\times\cdots\times T_pM}^{k+1\:\text{times}}$$ by extending this tuple to a $(k+1)-$tuple of smooth vector fields $X_1,\ldots, X_{k+1}$ so that $X_i(p)=v_i$. I can see that this is independent of extension, because of the corresponding fact for the coordinate definition.

On the other hand, I would like to show that this formula is well-defined in this sense intrinisically, based only on the coordinate independent formula.

EDIT: As an addendum, I want to understand why the coordinate independent definition is independent of extension $(X_1,\ldots, X_{k+1})$ chosen without appealing to the coordinate definition.

EDIT 2: Please note this question is not a duplicate of the other, because I am interested in showing the corresponding fact for the coordinate-independent version of the formula, without appealing to the coordinate expression.

• I only briefly scanned that question, but it does not address my question, since I am wondering why the coordinate free form is intrinsically well-defined, as opposed to the coordinate version, which I understand. – Antonios-Alexandros Robotis Nov 26 '17 at 21:51
• All right. Let me remove that. – Sou Nov 26 '17 at 21:55
• What do you mean by the last paragraph ? Its not really clear (at least for me). Can you make some sketch ? – Sou Nov 26 '17 at 22:08
• Possible duplicate of Exterior derivative well-defined – Leucippus Nov 27 '17 at 0:06
• As already addressed, that question does not answer my question. They are different questions, albeit related. – Antonios-Alexandros Robotis Nov 27 '17 at 0:07

If you're trying to avoid coordinates, the way to verify that an $\mathbb R$-multilinear expression involving vector fields is "pointwise" (i.e. depends only on the value of the vector fields at the point at which the expression itself is being evaluated) is via the following fact, sometimes called the Tensor Characterization Lemma.
A map $T:\mathscr X(M)\times\cdots \times \mathscr X(M) \to C^\infty(M)$ is a tensor field if and only if it is $C^\infty (M)-$multilinear.
Here I am using the pointwise/"smooth section" definition of tensor fields: a tensor field is a smooth map $s : M \to T^*M \otimes \ldots \otimes T^* M$ such that $\pi \circ s = \mathrm{id}_M$, where $\pi$ is the bundle projection sending a tensor to its base point. When we say $T$ is a tensor field, what we really mean is that $T$ can be written $$T(X_1,\ldots,X_n)(p) = s(p)(X_1|_p, \ldots, X_n|_p)$$ for some tensor field $s$. (In practice, since we have this equivalence we typically treat $s$ and $T$ as just being two different ways of looking at the same object, and many authors define tensor fields as $C^\infty(M)-$multilinear maps of vector fields instead.)
Thus to answer your question, it suffices to show that $(X_1,\ldots,X_n) \mapsto d \omega(X_1,\ldots,X_n)$ is $C^\infty(M)-$multilinear, which you have apparently already done.